Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2025 AMC 8 Problems/Problem 14"
 
(7 intermediate revisions by 4 users not shown)
Line 1: Line 1:
 +
==Problem==
 
A number <math>N</math> is inserted into the list <math>2</math>, <math>6</math>, <math>7</math>, <math>7</math>, <math>28</math>. The mean is now twice as great as the median. What is <math>N</math>?
 
A number <math>N</math> is inserted into the list <math>2</math>, <math>6</math>, <math>7</math>, <math>7</math>, <math>28</math>. The mean is now twice as great as the median. What is <math>N</math>?
  
Line 6: Line 7:
  
 
The median of the list is <math>7</math>, so the mean of the new list will be <math>7 \cdot 2 = 14</math>. Since there will be <math>6</math> numbers in the new list, the sum  of the <math>6</math> numbers will be <math>14 \cdot 6 = 84</math>. Therefore, <math>2+6+7+7+28+N = 84 \rightarrow N = \boxed{\text{(E)\ 34}}</math>
 
The median of the list is <math>7</math>, so the mean of the new list will be <math>7 \cdot 2 = 14</math>. Since there will be <math>6</math> numbers in the new list, the sum  of the <math>6</math> numbers will be <math>14 \cdot 6 = 84</math>. Therefore, <math>2+6+7+7+28+N = 84 \rightarrow N = \boxed{\text{(E)\ 34}}</math>
 +
 +
~Soupboy0
 +
 +
==Solution 2 (Using answer choices)==
 +
 +
We could use answer choices to solve this problem. The sum of the <math>5</math> numbers is <math>50</math>. If you add <math>7</math> to the list, <math>57</math> is not divisible by <math>6</math>, therefore it will not work. Same thing applies to <math>14</math> and <math>20</math>. The only possible choices left are <math>28</math> and <math>34</math>. You now check <math>28</math>. <math>28</math> doesn't work because <math>(28+50) \div 6 = 13</math> and <math>13</math> is not twice of the median, which is still <math>7</math>. Therefore, only choice left is <math>\boxed{\text{(E)\ 34}}</math>
 +
 +
~HydroMathGod
 +
 +
==Vide Solution 1 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=jTTcscvcQmI
 +
 +
==Video Solution by Thinking Feet==
 +
https://youtu.be/PKMpTS6b988
 +
 +
==See Also==
 +
{{AMC8 box|year=2025|num-b=13|num-a=15}}
 +
{{MAA Notice}}

Latest revision as of 19:06, 30 January 2025

Problem

A number $N$ is inserted into the list $2$, $6$, $7$, $7$, $28$. The mean is now twice as great as the median. What is $N$?

$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 14\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 28\qquad \textbf{(E)}\ 34$

Solution

The median of the list is $7$, so the mean of the new list will be $7 \cdot 2 = 14$. Since there will be $6$ numbers in the new list, the sum of the $6$ numbers will be $14 \cdot 6 = 84$. Therefore, $2+6+7+7+28+N = 84 \rightarrow N = \boxed{\text{(E)\ 34}}$

~Soupboy0

Solution 2 (Using answer choices)

We could use answer choices to solve this problem. The sum of the $5$ numbers is $50$. If you add $7$ to the list, $57$ is not divisible by $6$, therefore it will not work. Same thing applies to $14$ and $20$. The only possible choices left are $28$ and $34$. You now check $28$. $28$ doesn't work because $(28+50) \div 6 = 13$ and $13$ is not twice of the median, which is still $7$. Therefore, only choice left is $\boxed{\text{(E)\ 34}}$

~HydroMathGod

Vide Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

See Also

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AMC_8_Problems/Problem_14&oldid=241314"